Transcript Review2

Review - II (chapters 5 and 6)
Newton's 1st law: If no force acts on a body, then
the body's velocity cannot change; that is, it cannot
accelerate.
1 Newton is that force required to accelerate our
standardized mass (1 Kg) at a rate of 1 m.s-2.
Mass is simply the characteristic of a body that relates
a force on the body to the resulting acceleration
Fnet  ma
Newton's 2nd law:
F1
Free-body
diagrams
S
F3
F1
F2
F2
F3
F  0  a
N  Fg  ma
Normal force
N  Fg  ma y
N  ma y  Fg  m  a y  g 
N  Fg  mg if a y  0
Weight (a force!):
•In the above example, the internal forces within the table
supply the normal force, which is normal to the surface.
•If we hold the mass in a stationary state, we must supply
the force. This is the sensation of weight, i.e.
W  Fg  mg Newtons
Friction and tension
•We will deal with friction next week
(chapter 6). All you need to know that a
friction force acts parallel to a surface
in the opposite direction to the motion.
•A taut cord is said to be in a state
of tension.
•If the body pulling on the cord does
so with a force of 50 N, then the
tension in the cord is 50 N.
•A taut cord pulls on objects at either end with equal and
opposite force equal to the tension.
•Cords are massless, pulleys are massless and frictionless
Newton's 3rd law
When two bodies interact, the forces on the bodies
from each other are always equal in magnitude and
opposite in direction.
For every "action" force, there is always an equal
and opposite "reaction" force; we call these a
"third-law force pair."
•When a table supports an object against the force of
gravity, the internal forces within the table supply an
upward normal force, which is normal to the surface.
•If we hold the mass in a stationary state, we must supply
the normal force. This is the sensation of weight, i.e.
W  Fg  mg Newtons
Review of static friction
1.
In static situations, the static frictional force exactly
cancels the component of the applied force parallel to
the surface.
2. There is a maximum static frictional force which
depends on the normal force between the surface and
the object, i.e.
f s ,max   s N
where s is the coefficient of static friction and N is
the magnitude of the normal force. s is a parameter
that depends on both surfaces. Once the force
component parallel to the surface exceeds fs,max, then the
body begins to slide along the surface.
Review of kinetic friction
3. If a body begins to slide along the surface, the
magnitude of the frictional force instantly decreases to
a value fk given by
fk  k N
where k is the coefficient of kinetic friction and N is
the magnitude of the normal force. Therefore, during
the sliding, a kinetic frictional force of magnitude fk
opposes the motion.
4. When several agents push in different directions on an
object, the frictional force opposes the component of
the net force on the object which is parallel to the
surface.
Review of Drag force and terminal speed
D  C  Av
1
2
DMA
2
Newton's 2nd law:
Mass
D  F  C  Av  mg  ma
g
2
1
2
Terminal speed when a = 0.
1
2
g
ME
F
C  Av  mg
2
2mg
or v 
C A
Review of uniform circular motion
•Although v does not change, the
direction of the motion does, i.e. the
velocity (a vector) changes.
•Thus, there is an acceleration
associated with the motion.
•We call this a centripetal
acceleration.
v2
a
r
mv
F  ma 
r
2
2 r
Period: T 
v
•Since v does not change, the acceleration and force
must be perpendicular to the velocity, i.e. directed
towards the center of the motion.
Energy
Kinetic energy K is energy associated with the state
of motion of an object. The faster an object moves,
the greater its kinetic energy.
•Definition:
K  12 mv 2
•SI unit is the joule (J):
1 joule = 1 J = 1 kg.m2/s2
Work
Work W is the energy transferred to or from an
object by means of a force acting on the object.
Energy transferred to the object is positive work, and
energy transferred from the object is negative work.
Fx  F cos ;
W  Fd cos  ; W  F  d
Work done by a spring force
•Hooke's law:
F  kd
dis the displacement of the free end of the spring from its
position when in a relaxed, or equilibrium state.
k is the spring constant, or force constant, and is a measure
of the stiffness of the spring. It has dimensions of N.m-1.
•Hooke's law (scalar version):
 Ws  12 kxi2  12 kx 2f
F  kx
work done by the spring
Work is positive if the mass ends up closer to the
relaxed position than it was initially. It is negative if the
mass ends up further away.
If xi = 0 and we call the final position x, then Ws = ½kx2
Work done by an applied force
•This is very similar to the situation in which we do work
against gravity.
•If we do work against a spring, then we do work on the
spring, while the spring does work on us.
•If the end of the spring is stationary before and after a
displacement, then
K  K f  K i  Wa  Ws  0
 Wa  Ws
If a block that is attached to a spring is stationary
before and after a displacement, then the work done on
it by the displacing force is the negative of the work
done on it by the spring.
Power
•Power is defined as the "rate at which work is done."
•If an amount of work W is done in a time
interval t by a force, the average power
due to the force during the time interval
is defined as
•Instantaneous power is defined as
Pavg
W

t
 
dW d
dr
P

F F
 F v
dt
dt
dt
•The SI unit for power is the Watt (W).
1 watt = 1 W = 1 J/s = 0.738 ft · lb/s
1 horsepower = 1 hp = 550 ft · lb/s = 746 W
1 kilowatt-hour = 1 kW · h = (103 W)(3600 s) = 3.60 MJ
Energy conservation and potential energy
U  W
•Note: the change in potential energy is simply the
negative of the work done by the gravitational force.
•Therefore, we already know how to compute U.
Mechanical energy
•We define mechanical energy as the sum of the kinetic
and potential energy, i.e.
Emech  K  U
•This is true only for ideal systems in which the only
forces which act are what we call conservative forces.
•The only conservative forces which you will encounter in
PHY2048 are gravitational and spring forces.
Calculation of potential energy
xf
U  W    F ( x)dx
xi
Gravitational potential energy
yf
U    (mg )dy
yi
 mg  y f  yi   mg y
U  U  U i  mg  y  yi 
U  mgy
Elastic potential energy
xf
U    (kx)dx
xi
 kx  kx
1
2
2
f
2
i
1
2
U  U  0  kx  0
1
2
U  kx
1
2
2
2
Conservation of mechanical energy
In an isolated system where only conservative forces
cause energy changes, the kinetic energy and potential
energy can separately change, but their sum, the
mechanical energy of the system, cannot change.
Emech  U  K  0
When the mechanical energy of a system is conserved,
we can related the sum of kinetic and potential energy
at one instant to that at another instant without
consideration of the intermediate motion and without
finding the work done by the forces involved.
U 2  K2  U1  K1
Calculation of force from potential energy
xf
U  W    F ( x)dx
xi
•Working backwards, since U is related to F through an
integration, it should come as no surprise that F is
related to U through differentiation.
U ( x)  W   F ( x)x
Yielding:
U ( x)
F ( x)  
,
x
or
dU ( x)
F ( x)  
dx
Potential
energy
curve
Emech  K  U
• K cannot be
negative, since it
is proportional
to (velocity)2
•Therefore,
U  Emech
always!!
Equilibrium
points
F=0
Conservation of energy
The total energy of a system can change only by amounts of
energy W that are transferred to or from the system.
W  E  Emech  Eth
where Eth acknowledges the fact that mechanical energy
may be converted to thermal energy due to frictional
forces or air resistance
The total energy of an isolated system cannot change.
Emech  Eth  0
In an isolated system, we can relate the total energy at one
instant to the total energy at another instant without
considering the energies at intermediate times.
Systems of particles
m1 x1  m2 x2  m3 x3    mn xn
xcom 
M
1 n
  mi xi
M i 1
•Here, i is a running number, or index, that takes on
all integer values from 1 to n.
•In three-dimensions:
xcom
1

M
n
m x ;
i 1
i i
ycom
1

M
n
m y ;
i 1
i
i
zcom
rcom  xcomˆi  ycomˆj  zcom kˆ
1

M
n
m z
i 1
i i
Linear momentum
•Definition of linear momentum, p:
p  mv
•If one takes the derivative,
dp
dv
m
 ma  Fnet
dt
dt
The time rate of change of momentum of a particle is
equal to the net force acting on the particle and is in
the direction of the force.
Linear momentum of a system of particles
•A system of n particles has a total linear momentum
given by:
P  p1  p2  p3   pn
 mv1  mv2  mv3   mvn
 M vcom
The linear momentum of a system of particles is equal
to the product of the total mass M of the system and
the velocity of the center of mass.
dP
dvcom
M
 Ma  Fnet
dt
dt
Conservation of linear momentum
•For a system of n particles, if no net force acts on
the system:
P  constant
(closed, isolated system)
If no net external force acts on a system of particles,
the total linear momentum of the system cannot change
 total linear momentum   total linear momentum 



 at some initial time t   at some later time t

f
i 



•These are vector equations, i.e.
Px  constant; Py  constant; Pz  constant
If the component of the net external force on a closed
system is zero along an axis, then the component of the linear
momentum of the system along that axis cannot change.
Completely inelastic collision - general case
Before:
m1
v1i
v2i
m2
After:
m1 m2

m1v1i  m2v2i
vf 
m1  m2
v2i  0
Special case:

m1
vf 
v1i
m1  m2
vf
Perfect elastic collision - general case
Before:
m1
v1i
After:
General result:
m1  m2
2m2
v1 f 
v1i 
v2i
m1  m2
m1  m2
v2 f
2m1
m2  m1

v1i 
v2i
m1  m2
m1  m2
v2i
m2
m1
v1f
m2
v2f
If v2i  0
m1  m2
v1 f 
v1i
m1  m2
v2 f
2m1

v1i
m1  m2